Find two positive numbers whose product is 64 and the sum is minimum.

Answer

To find two positive numbers whose product is 64 and whose sum is minimized, we can follow these steps:

### Step 1: Define the variables
Let the two positive numbers be \( x \) and \( y \).

### Step 2: Set up the product condition
According to the problem, the product of these two numbers is given by:
\[ xy = 64 \]
From this equation, we can express \( y \) in terms of \( x \):
\[ y = \frac{64}{x} \]

### Step 3: Set up the sum function
We want to minimize the sum of these two numbers:
\[ S = x + y \]
Substituting \( y \) from the previous step, we have:
\[ S = x + \frac{64}{x} \]

### Step 4: Differentiate the sum function
To find the minimum sum, we need to differentiate \( S \) with respect to \( x \):
\[ \frac{dS}{dx} = 1 - \frac{64}{x^2} \]

### Step 5: Set the derivative to zero
To find the critical points, set the derivative equal to zero:
\[ 1 - \frac{64}{x^2} = 0 \]
Solving for \( x^2 \):
\[ \frac{64}{x^2} = 1 \implies x^2 = 64 \implies x = 8 \]

### Step 6: Find \( y \)
Now, substitute \( x = 8 \) back into the equation for \( y \):
\[ y = \frac{64}{8} = 8 \]

### Step 7: Conclusion
Thus, the two positive numbers that satisfy the conditions of the problem are:
\[ x = 8 \quad \text{and} \quad y = 8 \]

### Step 8: Verify the minimum
To ensure that this critical point gives a minimum, we can check the second derivative:
\[ \frac{d^2S}{dx^2} = \frac{128}{x^3} \]
Since \( x = 8 \) gives a positive value for the second derivative, it confirms that \( S \) is minimized at \( x = 8 \).

### Final Answer:
The two positive numbers are \( 8 \) and \( 8 \).

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